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Implementing mergesort in Ruby

NB I’ve been a mostly self-taught programmer since I was in middle school (about 20 years now). I’m in the processing of teaching myself some of the things I missed out on by not having a Computer Science education; I’m also new to Ruby. This post is me writing out my thoughts on this subject to better understand it for myself — maybe someone else will find it useful, but don’t mistake me for an expert.

Mergesort is a classic “divide and conquer” algorithm for sorting an array. We have a problem like this:

  • Input: An unsorted array of numbers, e.g. [1, 4, 3, 9].
  • Output: The input array, sorted from least to greatest, e.g. [1, 3, 4, 9].

Mergesort works by splitting the input array into progressively smaller sorted arrays and then merging those sorted arrays together to produce the final sorted array.

Here is an implementation of mergesort I wrote in Ruby. I’ll break it down in the rest of this post.

def mergesort(array)
  return array if array.size <= 1

  left = mergesort(array[0...array.length / 2])
  right = mergesort(array[array.length / 2..])
  merge(left, right)
end

def merge(left, right)
  sorted = []

  until left.empty? && right.empty? do
    if left[0].nil?
      sorted.push(right.shift) unless right[0].nil?
    elsif right[0].nil?
      sorted.push(left.shift)
    elsif left[0] <= right[0]
      sorted.push(left.shift)
    else
      sorted.push(right.shift)
    end
  end

  sorted
end

Mergesort is split into two functions: a function mergesort which splits the array into smaller pieces and calls the merge function on those smaller pieces, and another function, the merge function, which combines two already-sorted arrays into a single sorted array.

Let’s look at the merge function first.

def merge(left, right)
  sorted = []

  until left.empty? && right.empty? do
    if left[0].nil?
      sorted.push(right.shift) unless right[0].nil?
    elsif right[0].nil?
      sorted.push(left.shift)
    elsif left[0] <= right[0]
      sorted.push(left.shift)
    else
      sorted.push(right.shift)
    end
  end

  sorted
end

The merge function takes two arrays, left and right, as parameters. It goes through each array element by element comparing them — if the left array’s element is smaller, it takes that element from the left array and adds it to a new sorted array; then it compares the next left element with the same right element. It does this until both left and right arrays are empty, then returns the sorted array.

If I pass merge([9], [1]) it returns [1, 9].

Passing merge([0, 2], [1, 3]) returns [0, 1, 2, 3].

Passing merge([9, 1], [2, 3]), though, returns [2, 3, 9, 1] because merge isn’t a sorting algorithm on its own — it just merges together two already-sorted arrays. We need to pass merge already-sorted arrays in order for it to work.

The function mergesort does that by splitting the input array into successively smaller arrays, splitting the input array in half and recursively calling itself on the two halves until it reaches two arrays of size 11. Once it has two arrays of size 1, it calls merge on those two arrays and starts building up a single sorted array out of progressively larger sorted array inputs.

Adding some output to the program shows what’s going on a little more clearly.

def mergesort(array)
  return array if array.size <= 1
  puts "Splitting #{array} in two"
  left = mergesort(array[0...array.length / 2])
  right = mergesort(array[array.length / 2..])
  print "Merging #{left} and #{right} "
  merged = merge(left, right)
  puts "into #{merged}"
  merged
end

def merge(left, right)
  sorted = []
  until left.empty? && right.empty? do
    if left[0].nil?
      sorted.push(right.shift) unless right[0].nil?
    elsif right[0].nil?
      sorted.push(left.shift)
    elsif left[0] <= right[0]
      sorted.push(left.shift)
    else
      sorted.push(right.shift)
    end
  end
  sorted
end

unsorted_array = Array.new(4)
unsorted_array.map! { |x| x = rand(-100..100) }
puts "The unsorted array: #{unsorted_array}"
puts "The sorted array: #{mergesort(unsorted_array)}"

Running that outputs:

The unsorted array: [-75, 43, -70, -41]
Splitting [-75, 43, -70, -41] in two
Splitting [-75, 43] in two
Merging [-75] and [43] into [-75, 43]
Splitting [-70, -41] in two
Merging [-70] and [-41] into [-70, -41]
Merging [-75, 43] and [-70, -41] into [-75, -70, -41, 43]
The sorted array: [-75, -70, -41, 43]

You could visualize it like this:

img

I hope that makes things clear! I think I will save analyzing the runtime complexity2 or discussions of when to use mergesort versus other sorting algorithms as exercises for the reader, or maybe future blog posts.

Footnotes

1 Assuming an initial input array of size 2 or more — otherwise it would just return its input.

2 The diagram above would be a good starting point.